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Mathematics

Given A = [11−20] and B=[2−111]\begin{bmatrix}[r] 1 & 1 \ -2 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix}.

Solve for matrix X :

(i) X + 2A = B

(ii) 3x + B + 2A = 0

(iii) 3A - 2X = X - 2B.

Matrices

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Answer

(i) Given,

⇒X+2A=B⇒X=B−2A⇒X=[2−111]−2[11−20]⇒X=[2−111]−[22−40]⇒X=[2−2−1−21−(−4)1−0]⇒X=[0−351]\Rightarrow X + 2A = B \\[1em] \Rightarrow X = B - 2A \\[1em] \Rightarrow X = \begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix} - 2\begin{bmatrix}[r] 1 & 1 \ -2 & 0 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix} - \begin{bmatrix}[r] 2 & 2 \ -4 & 0 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] 2 - 2 & -1 - 2 \ 1 - (-4) & 1 - 0 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] 0 & -3 \ 5 & 1 \end{bmatrix}

Hence, X = [0−351].\begin{bmatrix}[r] 0 & -3 \ 5 & 1 \end{bmatrix}.

(ii) Given,

⇒3X+B+2A=0⇒3X=−(B+2A)⇒3X=−([2−111]+2[11−20])⇒3X=−([2−111]+[22−40])⇒3X=−([2+2−1+21+(−4)1+0])⇒3X=−([41−31])⇒X=−13([41−31])⇒X=[−43−131−13]\Rightarrow 3X + B + 2A = 0 \\[1em] \Rightarrow 3X = -(B + 2A) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix} + 2\begin{bmatrix}[r] 1 & 1 \ -2 & 0 \end{bmatrix}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix} + \begin{bmatrix}[r] 2 & 2 \ -4 & 0 \end{bmatrix}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix}[r] 2 + 2 & -1 + 2 \ 1 + (-4) & 1 + 0 \end{bmatrix}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix}[r] 4 & 1 \ -3 & 1 \end{bmatrix}\Big) \\[1em] \Rightarrow X = -\dfrac{1}{3}\Big(\begin{bmatrix}[r] 4 & 1 \ -3 & 1 \end{bmatrix}\Big) \\[1em] \Rightarrow X = \begin{bmatrix}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \ 1 & -\dfrac{1}{3} \end{bmatrix}

Hence, X = [−43−131−13]\begin{bmatrix}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \ 1 & -\dfrac{1}{3} \end{bmatrix}.

(iii) Given,

⇒3A−2X=X−2B⇒X+2X=3A+2B⇒3X=3A+2B⇒3X=3[11−20]+2[2−111]⇒3X=[33−60]+[4−222]⇒3X=[3+43+(−2)−6+20+2]⇒3X=[71−42]⇒X=13[71−42]⇒X=[7313−4323].\Rightarrow 3A - 2X = X - 2B \\[1em] \Rightarrow X + 2X = 3A + 2B \\[1em] \Rightarrow 3X = 3A + 2B \\[1em] \Rightarrow 3X = 3\begin{bmatrix}[r] 1 & 1 \ -2 & 0 \end{bmatrix} + 2\begin{bmatrix}[r] 2 & -1 \ 1 & 1 \end{bmatrix} \\[1em] \Rightarrow 3X = \begin{bmatrix}[r] 3 & 3 \ -6 & 0 \end{bmatrix} + \begin{bmatrix}[r] 4 & -2 \ 2 & 2 \end{bmatrix} \\[1em] \Rightarrow 3X = \begin{bmatrix}[r] 3 + 4 & 3 + (-2) \ -6 + 2 & 0 + 2 \end{bmatrix} \\[1em] \Rightarrow 3X = \begin{bmatrix}[r] 7 & 1 \ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{3}\begin{bmatrix}[r] 7 & 1 \ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] \dfrac{7}{3} & \dfrac{1}{3} \ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix}.

Hence, X = [7313−4323].\begin{bmatrix}[r] \dfrac{7}{3} & \dfrac{1}{3} \ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix}.

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